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I have some understanding problems with the open mapping theorem.

Lemma

Let $f\in H(D(z_{0},\ R))$ ($f$ analytical on the disc with the origin in $z_0$ and radius $R$) and $$ |f(z_{0})|<\min_{|z-z_{0}|=r}|f(z)| $$ for a $r\in(0,\ R)$. Then there is a $w\in D(z_{0},\ r)$ with $f(w)=0$.

Theorem

Let $\Omega\subseteq \mathbb{C}$ be a region and $f\in H(\Omega)$. Then we have either $f(\Omega)$ is also a region or $f$ is constant.

Proof

We choose $f$ not constant. $f(\Omega)$ is connected as a continuous image of a connected set. We show, that $f(\Omega)$ is open:

Let $w_{0}\in f(\Omega)$, so $w_{0}=f(z_{0})$ for a $ z_{0}\in\Omega$. There is a $\delta>0$ with $$ f(z)\neq w_{0}\ (z\in\partial D(z_{0},\delta)),\ (a) $$ because otherwise $f(z)=w_{0}$ for all $ z\in\Omega$.

$\color{green}{\text{I don't understand why there is a } \delta \text{ so that (a) is fulfilled. I don't understand the followup.}}$

$\color{green}{\text{What is actually the outline of proving the openness in the next steps?}}$ $\color{green}{\text{(Read: I have no idea what happens in the steps below)?}}$

Because $f$ is continuous, there is a $\varepsilon>0$ with $$ |f(z)-w_{0}|\geq 3\varepsilon\ (z\in\partial D(z_{0},\delta)). $$ We show: $D(w_{0},e)\subseteq f(\Omega)$: For $|w-w_{0}|<\varepsilon$ and $|z-z_{0}|=\delta$ we have: $$ |f(z)-w|=|f(z)-w_{0}+w_{0}-w|\geq|f(z)-w_{0}|-|w_{0}-w|\geq 3e-e=2e $$ and $|f(z_{0})-w|=|w_{0}-w|<\varepsilon$, so that $$ |f(z_{0})-w|<\min_{|z-z_{0}|=\delta}|f(z)-w|. $$ Using the Lemma presented before the theorem, we have that $f(z)-w$ has a zero $z_{w}$ with $|z_{w}-z_{0}|<\delta$, so that $f(z_{w})=w$ with $ z_{w}\in\Omega.\ \square $

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up vote 2 down vote accepted
  1. If $\forall \delta$ there exists $z\in \partial D(z_0,\delta)$ such that $f(z) = w_0$, we would have that $z_0$ is an accumulation point of $f^{-1}(w_0)$. But since $f-w_0$ is holomorphic its roots can only accumulate if $f - w_0 \equiv 0$. This would contradict the assumption that $f$ is non constant. (For a proof of the accumulation point fact, see e.g. Theorem 4.8 in Chapter 2 of Stein and Shakarchi's Complex Analysis.)

  2. The remainder of the proof is setting up to apply the Lemma, which is a corollary of the maximum principle. Now, consider the function $g(z) = f(z) - w_0$. This function takes 0 at $z_0$. By the previous step we see that along the boundary of some disk $D(z_0,\delta)$ $g \neq 0$, and so is bounded away from zero. So if we subtract from $g$ a sufficiently small number, $g(z_0) - \Delta w$ is still going to be much smaller than $g(z) - \Delta w$ along $\partial D(z_0,\delta)$, and we can apply the Lemma.

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Thank you. Got the book as well at your suggestion! –  Chris Jun 6 '12 at 23:38
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